How to define what is the elementary operation in an algorithm?

Question

I always thought that the elementary operation from an algorithm was the operation located inside the most inner loop. I found very little detail about this in books and online articles, maybe because it was supposed to be something trivial, but the few that cared to explain what should be the elementary operation in an algorithm, they always says that its the operation that is executed the most, that is, the one located inside the most inner loop.

And as such, in this algorithm:

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        while (true) {
            int N = in.nextInt();
            if (N == 0)
                break;

            long cost = 0;
            int[] houses = new int[N];
            for (int i = 0; i < N; ++i)
                houses[i] = in.nextInt();

            for (int i = 0; i < N - 1; ++i) {
                cost += Math.abs(houses[i]);
                houses[i + 1] += houses[i];
            }
            System.out.println(cost);
        }
    }
}

I said that the elementary operation was the assignment operation houses[i] = in.nextInt(); inside the first for, because it runs N times while the second for runs N-1 times.

My teacher is saying that this is incorrect and that the elementary operation in this algorithm is the operation inside the second for.

By any chances, there has exceptions where the elementary operation isn't the operation located inside the most inner loop, or she is wrong?

score 0 · Answer 1 · answered Feb 18 '16 at 20:03

0

For this program, the second for loop is not inside the first for loop.

This program takes N numbers as inputs for the houses array and the actual algorithm is only in the second for loop, where it keeps the running count of the cost of the first i items of the array in the ith index.

Your teacher is correct.

answered Feb 18 '16 at 20:03

Brendan

303
1
6

But if we consider the instructions inside the loops as costing O(1), it's like the operation inside the first for costs O(1) at each time it is executed and the operations inside the second for costs together O(2) at each time they are executed. But the cost of the for loops is O(n) and as O(1) and O(2) are irrelevant if compared with O(n), I thought that we should disconsider the codes inside the loops and just consider the cost of the loops itself... And by doing that, since the first for runs more times than the second for, I though that it would be considered the elementary operation. – Wartogh Feb 19 '16 at 21:39
@Brendan, it's not clear why elementary operation should be in the second for loop. Can you please elaborate ? – user9026 Sep 29 '18 at 11:30

score 0 · Answer 2 · answered May 20 '20 at 19:51

An elementary operation is one whose execution time is bounded by a constant for a particular machine and programming language. That means if we run an elementary operation written in a specific language on a specific machine, it always takes the same time to be executed. Such as assignment, addition, multiplication, etc. In your case, the operation in.nextInt(); under the first loop actually is a keyboard input event, kind of IO operations, which depends mainly on users. So it's not an elementary operation.

How to define what is the elementary operation in an algorithm?

2 Answers2